Question

Air at \(52^{\circ} \mathrm{C}, 101.3 \mathrm{kPa}\), and 20 percent relative humidity enters a \(5-\mathrm{cm}\)-diameter tube with an average velocity of $6 \mathrm{~m} / \mathrm{s}$. The tube inner surface is wetted uniformly with water, whose vapor pressure at \(52^{\circ} \mathrm{C}\) is \(13.6 \mathrm{kPa}\). While the temperature and pressure of air remain constant, the partial pressure of vapor in the outlet air is increased to \(10 \mathrm{kPa}\). Detemine \((a)\) the average mass transfer coefficient in $\mathrm{m} / \mathrm{s}\(, \)(b)$ the log-mean driving force for mass transfer in molar concentration units, \((c)\) the water evaporation rate in $\mathrm{kg} / \mathrm{h}\(, and \)(d)$ the length of the tube.

Short answer

Answer: The average mass transfer coefficient is \(4.15 \times 10^{-3}\,\mathrm{m/s}\), the log-mean driving force for mass transfer is \(0.0185 \,\mathrm{mol/m^3}\), the water evaporation rate is \(0.0103\,\mathrm{kg/h}\), and the length of the tube is \(34.1\,\mathrm{m}\).

Step-by-step solution

Calculate the input air molar concentration

At \(52^\circ\mathrm{C}, 101.3 \mathrm{kPa}\), the partial pressure of water vapor (\(P_v\)) at 20% relative humidity can be found using the given vapor pressure of water at \(52^\circ\mathrm{C}\). \(P_v = (0.2)(13.6 \mathrm{kPa})=2.72 \mathrm{kPa}\) We can find the molar concentration (\(C_v\)) of water vapor at the inlet, using the Ideal Gas Law: \(C_v = \frac{P_v}{RT} = \frac{2.72 \mathrm{kPa}}{(8.314 \frac{J}{\mathrm{mol}\cdot K})(52 + 273.15) \mathrm{K}} = 0.0101 \mathrm{mol/m^3}\)

Calculate the outlet air molar concentration

The partial pressure of vapor in the outlet air is given as \(10 \mathrm{kPa}\). We again use the Ideal Gas Law to find the molar concentration of water vapor at the outlet: \(C_{v_{out}} = \frac{10 \mathrm{kPa}}{(8.314 \frac{J}{\mathrm{mol}\cdot K})(52 + 273.15) \mathrm{K}} = 0.0372 \mathrm{mol/m^3}\)

Find the average mass transfer coefficient

We can now determine the average mass transfer coefficient (\(k_v\)) using the molar concentration difference between the inlet and outlet, and the partial pressure difference: \(k_v = \frac{C_{v_{out}} - C_v}{10 \mathrm{kPa} - 2.72 \mathrm{kPa}} = 4.15 \times 10^{-3} \mathrm{m/s}\)

Calculate the log-mean driving force

The log-mean driving force for mass transfer is given by the following formula: \(LDF = \frac{(C_{v_{out}} - C_v)}{\ln{(\frac{C_{v_{out}}}{C_v})}}\) \(LDF = \frac{0.0271 \,\mathrm{mol/m^3}}{\ln{(\frac{0.0372}{0.0101})}} = 0.0185 \,\mathrm{mol/m^3}\)

Determine the evaporation rate

The water evaporation rate can be found using the mass transfer coefficient and the log-mean driving force: \(Evaporation \, Rate = k_v \times LDF \times Area\) Observing the given diameter of the tube, we can calculate the area: \(Area = \pi (\frac{0.05\,\mathrm{m}}{2})^2 = 0.00196\,\mathrm{m^3}\) \(Evaporation \, Rate = (4.15 \times 10^{-3}\,\mathrm{m/s}) \times (0.0185\,\mathrm{mol/m^3}) \times (0.00196 \,\mathrm{+m^2}) = 1.58 \times 10^{-7}\,\mathrm{mol/s}\) To convert the evaporation rate into \(\mathrm{kg/h}\), we multiply by the molar mass of water: \(Evaporation \, Rate (kg/h) = (1.58 \times 10^{-7}\,\mathrm{mol/s}) \times (18.015\,\mathrm{g/mol}) \times (\frac{3600\,\mathrm{s}}{1\,\mathrm{h}})\times (\frac{1\,\mathrm{kg}}{1000\,\mathrm{g}})= 0.0103\,\mathrm{kg/h}\)

Calculate the length of the tube

We can now find the length of the tube using the evaporation rate and the mass transfer rate: \(Length = \frac{Evaporation \, Rate}{k_v \times LDF \times Velocity}\) \(Length = \frac{0.0103\,\mathrm{kg/h}}{(4.15 \times 10^{-3}\,\mathrm{m/s}) \times (0.0185\,\mathrm{mol/m^3}) \times (6\,\mathrm{m/s})}= 34.1\,\mathrm{m}\) So, the results are as follows: a) The average mass transfer coefficient is \(4.15 \times 10^{-3}\,\mathrm{m/s}\). b) The log-mean driving force for mass transfer is \(0.0185 \,\mathrm{mol/m^3}\). c) The water evaporation rate is \(0.0103\,\mathrm{kg/h}\). d) The length of the tube is \(34.1\,\mathrm{m}\).